Match the species in List I with their geometry in List II

List I	List II
A. PCl$_5$	I. Tetrahedral
B. BrF$_5$	II. Square Planar
C. BF$_4^-$	III. Trigonal bipyramidal
D. [Ni(CN)$_4$]$^{2-}$	IV. Square pyramidal

Choose the correct answer from the options given below:

For $\text{PCl}_5$, the central atom $\text{P}$ has $5$ valence electrons. It forms $5$ bond pairs with $\text{Cl}$ atoms and has $0$ lone pairs. The hybridization is $sp^3d$, and the geometry is trigonal bipyramidal. (A $\rightarrow$ III)

For $\text{BrF}_5$, the central atom $\text{Br}$ has $7$ valence electrons. It forms $5$ bond pairs with $\text{F}$ atoms and has $1$ lone pair. The hybridization is $sp^3d^2$, and the geometry is square pyramidal. (B $\rightarrow$ IV)

For $\text{BF}_4^-$, the central atom $\text{B}$ has $3$ valence electrons plus $1$ from the negative charge, giving $4$ electrons. It forms $4$ bond pairs with $\text{F}$ atoms and has $0$ lone pairs. The hybridization is $sp^3$, and the geometry is tetrahedral. (C $\rightarrow$ I)

For $[\text{Ni}(\text{CN})_4]^{2-}$, the central metal ion $\text{Ni}^{2+}$ has a $3d^8$ electronic configuration. Since $\text{CN}^-$ is a strong field ligand, it causes the pairing of electrons in the $3d$ orbitals. The hybridization is $dsp^2$, and the geometry is square planar. (D $\rightarrow$ II)

The correct matching is A-III, B-IV, C-I, D-II.

Answer: A-III, B-IV, C-I, D-II